Driven Oscillator: Amplitude, Phase and Resonance

Derivation

Driven Oscillator: Amplitude, Phase and Resonance

D-061 Home PU-103 Threads force · energy Depends on The Damped Harmonic Oscillator, complex-exponential-method

Statement

For a mass on a linear spring with viscous damping, driven by a sinusoidal force \(F_0\cos\omega t\), the steady-state motion is oscillation at the drive frequency \(\omega\) with amplitude \(A(\omega)=\dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\zeta\omega_0\omega)^2}}\) and phase lag \(\phi(\omega)=\arctan\!\dfrac{2\zeta\omega_0\omega}{\omega_0^2-\omega^2}\); the amplitude peaks at \(\omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\) while the phase sweeps monotonically from \(0\) through \(\pi/2\) at \(\omega=\omega_0\) to \(\pi\) as \(\omega\to\infty\).

Why it matters

Resonance is the mechanism by which a small periodic input produces a large response: a swing pushed at its natural rhythm, a tuned radio circuit selecting one carrier from many, a bridge or building responding to periodic loading. The amplitude and phase functions derived here are the universal templates for every linear resonant system, mechanical or electrical.

The phase lag is the physically subtle half of the result and the part students most often neglect. It encodes when the driven oscillator absorbs energy most efficiently and why the maximum energy transfer occurs exactly at \(\omega_0\) rather than at the amplitude peak. The two frequencies differ, and the difference is the whole story of dissipative resonance.

Assumptions

Linear restoring force and linear (viscous) damping.If the spring stiffens or the damping goes as \(v^2\), the equation becomes nonlinear; superposition fails, the response is no longer monochromatic, and features like amplitude-dependent frequency and jump hysteresis appear (Duffing regime).

Steady state: transients have decayed.Immediately after switch-on the full solution contains a homogeneous part oscillating near \(\omega_0\) and decaying as \(e^{-\zeta\omega_0 t}\). Drop this assumption and you must add beats between drive and natural motion; the clean single-frequency response holds only for \(t\gg 1/(\zeta\omega_0)\).

Constant coefficients \(m,\gamma,k\).If parameters drift (heating changes \(\gamma\), a softening spring changes \(k\)) the resonance curve shifts in time and the phasor treatment no longer applies.

Underdamped system, \(\zeta<1/\sqrt2\).The amplitude resonance peak at \(\omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\) exists as a real frequency only for \(\zeta<1/\sqrt2\). For heavier damping the amplitude is monotonically decreasing in \(\omega\) and there is no peak, though the phase still passes through \(\pi/2\) at \(\omega_0\).

Derivation

\[ m\ddot{x} + \gamma\dot{x} + kx = F_0\cos\omega t \]

Newton's second law for the mass: inertial term, viscous drag \(-\gamma\dot x\), Hooke restoring force \(-kx\), and the external drive. A

\[ \ddot{x} + 2\zeta\omega_0\,\dot{x} + \omega_0^2\,x = \frac{F_0}{m}\cos\omega t, \qquad \omega_0^2=\frac{k}{m},\ \ 2\zeta\omega_0=\frac{\gamma}{m} \]

Divide by \(m\) and introduce the natural frequency \(\omega_0\) and dimensionless damping ratio \(\zeta\); this is the canonical form shared with the free damped oscillator. A

\[ \ddot{z} + 2\zeta\omega_0\,\dot{z} + \omega_0^2\,z = \frac{F_0}{m}\,e^{i\omega t}, \qquad x=\operatorname{Re} z \]

Complexify the drive using \(\cos\omega t=\operatorname{Re}\,e^{i\omega t}\). Because the operator has real coefficients, the real part of the complex solution solves the real equation. B

\[ z(t) = Z\,e^{i\omega t}, \qquad Z\in\mathbb{C} \]

Steady-state ansatz: the system responds at the drive frequency (the homogeneous solution has decayed). \(Z\) is a complex amplitude carrying both magnitude and phase. B

\[ \big(-\omega^2 + 2i\zeta\omega_0\omega + \omega_0^2\big)Z\,e^{i\omega t} = \frac{F_0}{m}\,e^{i\omega t} \]

Substitute the ansatz; each time derivative brings a factor \(i\omega\). The differential equation collapses to algebra because \(e^{i\omega t}\) is an eigenfunction of \(d/dt\). A

\[ Z = \frac{F_0/m}{(\omega_0^2-\omega^2) + 2i\zeta\omega_0\omega} \]

Cancel the common \(e^{i\omega t}\) and solve for \(Z\). This complex transfer amplitude contains the complete steady-state response. A

\[ Z = |Z|\,e^{-i\phi}, \quad |Z|=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\zeta\omega_0\omega)^2}}, \quad \phi=\arg\big[(\omega_0^2-\omega^2)+2i\zeta\omega_0\omega\big] \]

Write the complex quotient in polar form. The magnitude is \(|\text{numerator}|/|\text{denominator}|\); the phase of \(Z\) is minus the phase of the denominator, so the response lags the drive by \(\phi\). B

\[ x(t)=\operatorname{Re}\big(|Z|e^{-i\phi}e^{i\omega t}\big) = A(\omega)\cos(\omega t-\phi) \]

Take the real part. The physical displacement is a cosine at the drive frequency, delayed by phase \(\phi\), with amplitude \(A=|Z|\). A

\[ A(\omega)=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\zeta\omega_0\omega)^2}}, \qquad \tan\phi(\omega)=\frac{2\zeta\omega_0\omega}{\omega_0^2-\omega^2} \]

Read off amplitude and phase from step 7. The branch of \(\arctan\) is chosen so \(\phi\) runs continuously from \(0\) to \(\pi\): the denominator changes sign at \(\omega=\omega_0\), where \(\phi=\pi/2\). B

\[ \frac{d}{d\omega}\Big[(\omega_0^2-\omega^2)^2+(2\zeta\omega_0\omega)^2\Big]=0 \;\Rightarrow\; -4\omega(\omega_0^2-\omega^2)+8\zeta^2\omega_0^2\omega=0 \]

Amplitude is maximal where the denominator (the radicand) is minimal. Differentiate and set to zero; the trivial root \(\omega=0\) is discarded. C

\[ \omega^2=\omega_0^2-2\zeta^2\omega_0^2 \;\Rightarrow\; \boxed{\ \omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\ } \]

Solve for the peak. The amplitude resonance is pulled below \(\omega_0\) by damping; it is real only for \(\zeta<1/\sqrt2\), and equals \(\omega_0\) exactly only in the zero-damping limit. C

Result

\[ x(t)=A(\omega)\cos(\omega t-\phi),\quad A(\omega)=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\zeta\omega_0\omega)^2}},\quad \phi(\omega)=\arctan\frac{2\zeta\omega_0\omega}{\omega_0^2-\omega^2} \]

Reading. The mass settles into oscillation at the driving frequency, not its own natural frequency. The amplitude is large only when the drive is near \(\omega_0\), the sharpness of that peak set by the damping. The motion lags the force: in phase at low frequency (the spring can keep up), a quarter cycle behind at \(\omega_0\) (velocity in phase with force, maximum power absorption), and half a cycle behind at high frequency (inertia-dominated, mass moves opposite to the push). The amplitude peak \(\omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\) sits slightly below \(\omega_0\), whereas the peak of absorbed power sits exactly at \(\omega_0\).

Units check. \(F_0/m\) has units \(\mathrm{N\,kg^{-1}}=\mathrm{m\,s^{-2}}\). The denominator terms \((\omega_0^2-\omega^2)\) and \(2\zeta\omega_0\omega\) each have units \(\mathrm{s^{-2}}\), so the square root is \(\mathrm{s^{-2}}\). Thus \(A\) has units \(\mathrm{m\,s^{-2}}/\mathrm{s^{-2}}=\mathrm{m}\), a length. The phase \(\phi\) is the arctangent of a ratio of two \(\mathrm{s^{-2}}\) quantities, hence dimensionless (radians). Correct.

Limiting cases

Static limit \(\omega\to 0\): \(A\to F_0/(m\omega_0^2)=F_0/k\), the static Hooke's-law extension, and \(\phi\to 0\). The mass follows the force instantaneously.
High frequency \(\omega\to\infty\): \(A\to F_0/(m\omega^2)\to 0\) and \(\phi\to\pi\). The response is inertia-limited and exactly out of phase; the drive fights pure mass.
On natural frequency \(\omega=\omega_0\): \(A=F_0/(2m\zeta\omega_0^2)=F_0/(\gamma\omega_0)\) and \(\phi=\pi/2\) exactly, independent of the amplitude formula's peak location.
Weak damping \(\zeta\to 0\): \(\omega_{\text{res}}\to\omega_0\), the peak height \(A_{\max}\to F_0/(2m\zeta\omega_0^2)\to\infty\), and the phase transition sharpens toward a step at \(\omega_0\). Quality factor \(Q=1/(2\zeta)\to\infty\).
Critical/heavy damping \(\zeta\ge 1/\sqrt2\): no amplitude peak; \(A(\omega)\) decreases monotonically from \(F_0/k\). The phase still passes through \(\pi/2\) at \(\omega_0\).

Breaks when

Transient not yet decayed. For \(t\lesssim 1/(\zeta\omega_0)\) the homogeneous solution \(e^{-\zeta\omega_0 t}\cos(\omega_d t+\psi)\) is still present; the true motion beats between \(\omega\) and \(\omega_d=\omega_0\sqrt{1-\zeta^2}\) and no single amplitude or phase describes it.
Nonlinear restoring force or damping. Large amplitudes probe spring anharmonicity (Duffing) or drag becomes \(\propto v^2\); the resonance curve bends, becomes amplitude-dependent, and can show hysteretic jumps as \(\omega\) is swept up versus down. Superposition and the single-frequency response fail.
Zero damping exactly at \(\omega=\omega_0\). With \(\zeta=0\) the steady-state formula gives infinite amplitude; the true solution grows secularly as \(x\propto t\sin\omega_0 t\) and never reaches steady state. The bounded-response derivation is invalid.
Parametric or multi-frequency forcing. If the drive modulates a system parameter (e.g. \(\omega_0(t)\)) rather than adding a force, or contains several frequencies with nonlinearity present, the transfer-function picture breaks and instabilities or combination tones arise.

Failure modes

Confusing the three special frequencies. Students conflate the natural frequency \(\omega_0\), the damped free frequency \(\omega_d=\omega_0\sqrt{1-\zeta^2}\), and the amplitude-resonance frequency \(\omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\). All three coincide only as \(\zeta\to 0\); \(\omega_{\text{res}}<\omega_d<\omega_0\) in general.
Locating maximum power at the amplitude peak. Average absorbed power peaks at \(\omega=\omega_0\) exactly (velocity resonance), not at \(\omega_{\text{res}}\). The energy argument and the amplitude argument give different frequencies.
Dropping the phase. Taking \(x=A\cos\omega t\) instead of \(A\cos(\omega t-\phi)\). The phase lag is essential for energetics: power \(=F\dot x\) averages to \(\tfrac12 F_0\omega A\sin\phi\), which vanishes if you forget \(\phi\).
Wrong arctan branch. Using the principal value \(\arctan\) blindly gives a negative or discontinuous phase above \(\omega_0\), because \((\omega_0^2-\omega^2)\) turns negative. The correct \(\phi\) runs continuously \(0\to\pi\).
Including the transient in "steady state." Applying the amplitude formula at \(t=0\) right after switch-on, ignoring that it describes only the asymptotic motion.
Treating \(A\) as the initial displacement. \(A\) is the eventual steady oscillation amplitude, set by the balance of drive and dissipation, not by initial conditions (which affect only the decaying transient).

Discussion

The physical heart of the result is the interplay of three restoring influences competing with the drive: the spring \(kx\), the inertia \(m\ddot x\), and the damping \(\gamma\dot x\). At low frequency the spring dominates and the mass tracks the force in phase. At high frequency inertia dominates and the mass, unable to reverse fast enough, moves in antiphase. Exactly where spring and inertia cancel, \(\omega=\omega_0\), only the damping term opposes the drive; the response is limited purely by dissipation and the velocity is in phase with the force. This is why \(\phi=\pi/2\) at \(\omega_0\) is a robust, damping-independent landmark while the amplitude peak is not.

The distinction between the amplitude-resonance frequency \(\omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\) and the power-resonance frequency \(\omega_0\) is the signature of dissipative resonance. Amplitude asks "how far does the mass swing," a question the stored (potential) energy answers, and damping shifts its optimum downward. Power asks "how much energy per second does the drive feed in," answered by the velocity, whose amplitude \(\omega A\) peaks exactly at \(\omega_0\). Since the average input power \(\bar P=\tfrac12 F_0\omega A\sin\phi\) matches the average dissipated power \(\tfrac12\gamma\omega^2A^2\) in steady state, resonance is the frequency of maximal energy throughput, tying this page directly to the energy thread.

The width of the resonance curve is governed by the quality factor \(Q=\omega_0/(2\zeta\omega_0)=1/(2\zeta)\). The full width at half-maximum power is \(\Delta\omega\approx\omega_0/Q=2\zeta\omega_0\): a high-\(Q\) oscillator is sharply selective, ringing for \(\sim Q\) cycles after the drive stops. This single number connects the frequency-domain sharpness, the time-domain ringdown \(e^{-\omega_0 t/2Q}\), and the fractional energy lost per radian \(1/Q\).

Viewed as a linear filter, the complex amplitude \(Z(\omega)=\chi(\omega)F_0/m\) with susceptibility \(\chi(\omega)=[(\omega_0^2-\omega^2)+2i\zeta\omega_0\omega]^{-1}\) is the frequency-domain Green's function of the oscillator. Its poles sit in the lower half complex-\(\omega\) plane at \(\omega=-i\zeta\omega_0\pm\omega_0\sqrt{1-\zeta^2}\), whose imaginary parts are the transient decay rate and whose real parts are the ringing frequency. Causality of the response forces these poles into the lower half plane and, via contour integration, links the real (dispersive) and imaginary (absorptive) parts of \(\chi\) through the Kramers-Kronig relations. The Lorentzian absorption peak \(\operatorname{Im}\chi\) near \(\omega_0\) is the same lineshape seen in atomic spectra, driven LCR circuits, and NMR, where \(\zeta\) is replaced by the relevant relaxation rate.

Common misconceptions. Resonance is not "the frequency where the system naturally oscillates" made loud; it is where the drive most efficiently overcomes the combined impedance. Nor does maximum amplitude mean maximum energy input, the two frequencies differ. And infinite amplitude is a fiction of the undamped idealization: every real system has \(\zeta>0\), capping \(A\) at \(F_0/(\gamma\omega_0)\).

Worked examples

Example 1 — Amplitude and phase off resonance. A mass \(m=0.50\ \mathrm{kg}\) on a spring \(k=200\ \mathrm{N\,m^{-1}}\) has damping coefficient \(\gamma=2.0\ \mathrm{kg\,s^{-1}}\), driven by \(F_0=5.0\ \mathrm{N}\) at \(\omega=25\ \mathrm{rad\,s^{-1}}\). Find the steady-state amplitude and phase lag. B

\[ \omega_0=\sqrt{k/m}=\sqrt{200/0.50}=20\ \mathrm{rad\,s^{-1}}, \quad 2\zeta\omega_0=\gamma/m=2.0/0.50=4.0\ \mathrm{s^{-1}} \]

Compute natural frequency and the damping rate; here \(\zeta=4.0/(2\cdot20)=0.10\), underdamped. A

\[ \omega_0^2-\omega^2 = 400-625=-225\ \mathrm{s^{-2}}, \qquad 2\zeta\omega_0\omega = 4.0\times25 = 100\ \mathrm{s^{-2}} \]

Evaluate the two denominator terms; the first is negative because \(\omega>\omega_0\). A

\[ A=\frac{F_0/m}{\sqrt{(-225)^2+100^2}}=\frac{5.0/0.50}{\sqrt{50625+10000}}=\frac{10}{246.2}=0.0406\ \mathrm{m} \]

Insert into the amplitude formula; \(F_0/m=10\ \mathrm{m\,s^{-2}}\). A

\[ \phi=\arctan\frac{100}{-225}=\pi-\arctan\frac{100}{225}=\pi-0.418=2.72\ \mathrm{rad}\;(156^\circ) \]

Since \(\omega>\omega_0\) the denominator is negative, placing \(\phi\) in the second quadrant, past \(\pi/2\). B

\[ A\approx 4.1\ \mathrm{cm}, \qquad \phi\approx 2.72\ \mathrm{rad}=156^\circ \]

Reading. Driven above resonance, the amplitude is modest and the motion lags by more than a quarter cycle, heading toward antiphase.

Units check. \(A\) in metres, \(\phi\) dimensionless. Consistent.

Example 2 — Resonant peak location and height, and \(Q\). For the same system (\(\omega_0=20\ \mathrm{rad\,s^{-1}}\), \(\zeta=0.10\), \(F_0/m=10\ \mathrm{m\,s^{-2}}\)), find the amplitude-resonance frequency, the peak amplitude, and the quality factor. B

\[ \omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}=20\sqrt{1-2(0.10)^2}=20\sqrt{0.98}=19.80\ \mathrm{rad\,s^{-1}} \]

Apply the peak formula; the shift below \(\omega_0\) is only \(0.2\%\) for light damping. B

\[ (\omega_0^2-\omega_{\text{res}}^2)=400-392=8\ \mathrm{s^{-2}}, \quad 2\zeta\omega_0\omega_{\text{res}}=4.0\times19.80=79.2\ \mathrm{s^{-2}} \]

Evaluate the denominator terms at the peak frequency. A

\[ A_{\max}=\frac{10}{\sqrt{8^2+79.2^2}}=\frac{10}{\sqrt{64+6273}}=\frac{10}{79.6}=0.1256\ \mathrm{m} \]

Insert into the amplitude formula; compare the on-resonance estimate \(F_0/(\gamma\omega_0)=5.0/(2.0\times20)=0.125\ \mathrm{m}\), essentially identical for small \(\zeta\). B

\[ Q=\frac{1}{2\zeta}=\frac{1}{2(0.10)}=5.0, \qquad \Delta\omega\approx\frac{\omega_0}{Q}=\frac{20}{5.0}=4.0\ \mathrm{rad\,s^{-1}} \]

The quality factor sets the fractional width of the resonance and the number of ring-down cycles. B

\[ \omega_{\text{res}}\approx 19.8\ \mathrm{rad\,s^{-1}},\quad A_{\max}\approx 12.6\ \mathrm{cm},\quad Q=5.0 \]

Reading. The peak amplitude (12.6 cm) is about three times the off-resonance value from Example 1 (4.1 cm), and the modest \(Q=5\) means a fairly broad resonance that rings for roughly five cycles.

Units check. \(\omega_{\text{res}}\) in \(\mathrm{rad\,s^{-1}}\), \(A_{\max}\) in metres, \(Q\) dimensionless. Consistent.

Problems

An oscillator has \(\omega_0=50\ \mathrm{rad\,s^{-1}}\) and \(\zeta=0.05\). At what drive frequency is the phase lag exactly \(\pi/2\), and what is the phase lag at \(\omega=50\ \mathrm{rad\,s^{-1}}\)?
Solution
The phase lag is \(\pi/2\) precisely where \(\omega_0^2-\omega^2=0\), i.e. \(\omega=\omega_0=50\ \mathrm{rad\,s^{-1}}\), independent of \(\zeta\). Hence at \(\omega=50\ \mathrm{rad\,s^{-1}}\), \(\phi=\pi/2\) exactly (\(90^\circ\)). This is the damping-independent landmark: spring and inertia cancel, leaving only viscous opposition, so the velocity is in phase with the force.
For \(m=1.0\ \mathrm{kg}\), \(k=100\ \mathrm{N\,m^{-1}}\), \(\gamma=4.0\ \mathrm{kg\,s^{-1}}\), \(F_0=8.0\ \mathrm{N}\), find the static ( \(\omega\to0\) ) amplitude and the on-resonance amplitude at \(\omega=\omega_0\).
Solution
\(\omega_0=\sqrt{100/1.0}=10\ \mathrm{rad\,s^{-1}}\). Static: \(A(0)=F_0/k=8.0/100=0.080\ \mathrm{m}=8.0\ \mathrm{cm}\). On resonance: \(A(\omega_0)=F_0/(\gamma\omega_0)=8.0/(4.0\times10)=0.20\ \mathrm{m}=20\ \mathrm{cm}\). The resonant amplification factor is \(A(\omega_0)/A(0)=Q=1/(2\zeta)\). Here \(\zeta=\gamma/(2m\omega_0)=4.0/20=0.20\), so \(Q=2.5\), consistent with \(20/8.0=2.5\).
Show that the average power delivered by the drive in steady state is \(\bar P=\tfrac12 F_0\omega A\sin\phi\), and evaluate it at \(\omega=\omega_0\) for the system of Problem 2.
Solution
With \(x=A\cos(\omega t-\phi)\), \(\dot x=-A\omega\sin(\omega t-\phi)\) and \(F=F_0\cos\omega t\). Instantaneous power \(P=F\dot x=-F_0 A\omega\cos\omega t\sin(\omega t-\phi)\). Expand \(\sin(\omega t-\phi)=\sin\omega t\cos\phi-\cos\omega t\sin\phi\); over a cycle \(\langle\cos\omega t\sin\omega t\rangle=0\) and \(\langle\cos^2\omega t\rangle=\tfrac12\), giving \(\bar P=\tfrac12 F_0 A\omega\sin\phi\). At \(\omega=\omega_0\): \(\phi=\pi/2\) so \(\sin\phi=1\), \(A=0.20\ \mathrm{m}\), \(\omega_0=10\ \mathrm{rad\,s^{-1}}\), \(F_0=8.0\ \mathrm{N}\): \(\bar P=\tfrac12(8.0)(0.20)(10)(1)=8.0\ \mathrm{W}\). Check against dissipation \(\tfrac12\gamma\omega^2A^2=\tfrac12(4.0)(100)(0.04)=8.0\ \mathrm{W}\). They balance, as steady state requires.
A system has \(\omega_0=1000\ \mathrm{rad\,s^{-1}}\) and quality factor \(Q=200\). Estimate (a) the damping ratio \(\zeta\), (b) the full width \(\Delta\omega\) of the power resonance, and (c) the fractional downward shift of the amplitude peak from \(\omega_0\).
Solution
(a) \(Q=1/(2\zeta)\Rightarrow\zeta=1/(2Q)=1/400=2.5\times10^{-3}\). (b) \(\Delta\omega\approx\omega_0/Q=1000/200=5.0\ \mathrm{rad\,s^{-1}}\). (c) \(\omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\); the fractional shift \(\approx\zeta^2=6.25\times10^{-6}\), i.e. \(\omega_0-\omega_{\text{res}}\approx\zeta^2\omega_0\approx6.3\times10^{-3}\ \mathrm{rad\,s^{-1}}\). For a high-\(Q\) resonator the peak shift is utterly negligible while the width is what one measures.
An underdamped oscillator (\(\omega_0=8.0\ \mathrm{rad\,s^{-1}}\), \(\zeta=0.25\)) is driven at \(\omega=6.0\ \mathrm{rad\,s^{-1}}\) with \(F_0/m=3.0\ \mathrm{m\,s^{-2}}\). Find \(A\) and \(\phi\), and state whether the drive is above or below the amplitude-resonance frequency.
Solution
\(\omega_0^2-\omega^2=64-36=28\ \mathrm{s^{-2}}\); \(2\zeta\omega_0\omega=2(0.25)(8.0)(6.0)=24\ \mathrm{s^{-2}}\). \(A=\dfrac{3.0}{\sqrt{28^2+24^2}}=\dfrac{3.0}{\sqrt{784+576}}=\dfrac{3.0}{36.9}=0.0813\ \mathrm{m}\approx8.1\ \mathrm{cm}\). \(\phi=\arctan(24/28)=\arctan(0.857)=0.708\ \mathrm{rad}\approx40.6^\circ\); denominator positive so first quadrant, \(\phi<\pi/2\). Amplitude peak: \(\omega_{\text{res}}=8.0\sqrt{1-2(0.25)^2}=8.0\sqrt{0.875}=7.48\ \mathrm{rad\,s^{-1}}\). Since \(6.0<7.48\), the drive is below the amplitude-resonance frequency, consistent with \(\phi<\pi/2\).